# Derivation of Fokker Planck forward equation

I am studing the derivation of the Fokker-Planck equation. My source is this link http://www.pnas.org/content/pnas/suppl/2007/03/16/0608270104.DC1/08270Appendix.pdf. it is very clear, but I am struggling with one passage: I have the equation $$\begin{equation} \int h(y)\frac{\partial p(y, t|x, 0)}{\partial t}dy- \int p(z, t|x,0)\sum_{n=1}^{\infty}D_{(n)}(z)h^{(n)}(z)dz=0. \end{equation}$$ and I have to obtain, integrating by parts, $$\begin{equation} \int h(z)\left(\frac{\partial p(y, t|x, 0)}{\partial t} -\sum_{n=1}^{\infty}\left[\left(-\frac{\partial}{\partial z}\right)^n \left(D_{(n)} p(z,t|x,0)\right)\right] \right) dz = 0. \end{equation}$$ Could you please explain me what does the author intend with integrating by parts? How can I delete the extra parts that come out integrating by parts? Does that come from that fact that $$p$$ is a probability distribution so it is very small at infinity?

Thank you!

By integrating by parts $$n$$ times the second term, you are essentially transferring the $$n$$ derivatives from the term $$h^{(n)}(z)$$ to the term $$D_{(n)} p(z,t|x,0)$$, obtaining $$\left(-\frac{\partial}{\partial z}\right)^n D_{(n)} p(z,t|x,0)$$. The minus terms comes from the formula for integration by parts, $$\int_a^b u dv=uv\Big|_a^b-\int_a^b v du$$ (notice how the new integral on the right hand side is multiplied by $$-1$$ compared to the integral on the left hand side). Doing this $$n$$ times will give you a factor of $$(-1)^n$$, in addition to the $$n$$ derivatives from $$\left(\frac{\partial}{\partial z}\right)^n$$ because of the derivative transfer from $$h^{(n)}(z)$$ to $$D_{(n)} p(z,t|x,0)$$.
Now you may be wondering why the author ignores the extra $$h^{(n-k)}(z)\left(-\frac{\partial}{\partial z}\right)^k D_{(n)} p(z,t|x,0)\Biggr|_{-\infty}^\infty$$ term, for $$k=0,1,...n$$, when using integration by parts. This is because the function $$h$$ is defined to be smooth and have compact support by the author, between equations 4 and 5. As a result, that term will always vanish.
I think that the author integrated by parts $$n$$ times like this in order to be able to factor out the $$h(z)$$ function from both terms, to show that, because $$h(z)$$ is arbitrary, the other term must always be $$0$$: $$\frac{\partial p(y,t|x,0)}{\partial t}-\sum_{n=1}^\infty \left(-\frac{\partial}{\partial z}\right)^n D_{(n)} p(z,t|x,0)=0$$
$$\frac{\partial p(y,t|x,0)}{\partial t}=\sum_{n=1}^\infty \left(-\frac{\partial}{\partial z}\right)^n D_{(n)} p(z,t|x,0)$$